[题解] LuoguP2257 YY的GCD

传送门

给\(n,m\)，让你求

\[\sum\limits_{i=1}^n \sum\limits_{j=1}^m [\gcd(i,j) \in prime]
\]

有\(T\)组询问\((T \le 10^4,n,m\le 10^7)\)。

枚举质数\(p\)，然后柿子变成

\[\sum\limits_{p} \sum\limits_{i=1}^n\sum\limits_{j=1}^m [\gcd(i,j)=p]
\]

等价于

\[\sum\limits_{p} \sum\limits_{i=1}^{\left\lfloor n/p\right\rfloor}\sum\limits_{j=1}^{\left\lfloor m/p\right\rfloor} [\gcd(i,j)=1]
\]

因为把\(\left\lfloor n/p \right\rfloor,\left\lfloor m/p\right\rfloor\)以内互质的一对数乘上\(p\)就是\(\gcd=p\)的数了。

令\(S(n,m)=\sum\limits_{i=1}^n\sum\limits_{j=1}^m [\gcd(i,j)=1]\)，由于\(\mu\)的性质\([n=1]=\sum\limits_{d\mid n} \mu(d)\)，所以

\[\begin{aligned}S(n,m)&=\sum\limits_{i=1}^n \sum\limits_{j=1}^m \sum\limits_{d\mid \gcd} \mu(d) \\&= \sum\limits_{d}\mu(d)\sum\limits_{i=1}^n [d\mid i]\sum\limits_{j=1}^m [d\mid j] \\&= \sum\limits_{d} \mu(d)\left\lfloor \frac{n}{d}\right\rfloor\left\lfloor \frac{m}{d}\right\rfloor\end{aligned}
\]

又因为\(\left\lfloor\frac{\left\lfloor\frac{n}{p}\right\rfloor}{d}\right\rfloor=\left\lfloor\frac{n}{dp}\right\rfloor\)，所以柿子是

\[\sum\limits_{p} \sum\limits_{d} \mu(d) \left\lfloor\frac{n}{dp}\right\rfloor \left\lfloor\frac{m}{dp}\right\rfloor
\]

可以枚举\(k=dp\)，有

\[\sum\limits_{k} \left\lfloor\frac{n}{k}\right\rfloor \left\lfloor\frac{m}{k}\right\rfloor\sum\limits_{p\mid k} \mu\left(\frac{k}{p}\right)
\]

令\(f(n)=\sum\limits_{p\mid n} \mu(\frac{n}{p})\)，预处理出\(f\)的前缀和，数论分块就好了。

并不会算预处理的复杂度qwq...

#include <bits/stdc++.h>
using namespace std;
#define rep(i,a,b) for (int i=(a);i<(b);++i)
#define per(i,a,b) for (int i=(a)-1;i>=(b);--i)
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define all(x) (x).begin(),(x).end()
#define SZ(x) ((int)(x).size())
typedef double db;
typedef long long ll;
typedef pair<int,int> PII;
typedef vector<int> VI;

const int maxn=1e7,N=maxn+10;
int vis[N],p[N],pn,mu[N],sum[N];

#define ss(l,r) (sum[r]-sum[l-1])

void init(int n) {
    mu[1]=1;
    rep(i,2,n+1) {
        if(!vis[i]) {p[pn++]=i;mu[i]=-1;}
        for(int j=0;j<pn&&i*p[j]<=n;j++) {
            vis[i*p[j]]=1;
            if(i%p[j]==0) {mu[i*p[j]]=0;break;}
            else mu[i*p[j]]=-mu[i];
        }
    }
    rep(i,0,pn) for(int j=p[i];j<=n;j+=p[i])
        sum[j]+=mu[j/p[i]];
    rep(i,1,n+1) sum[i]+=sum[i-1];
}

ll solve(int n,int m) {
    int tn=min(n,m); ll ans=0;
    for(int l=1,r=0;l<=tn;l=r+1) {
        r=min(n/(n/l),m/(m/l));
        ans+=1ll*(n/l)*(m/l)*ss(l,r);
    }
    return ans;
}

int main() {
#ifdef LOCAL
    freopen("a.in","r",stdin);
#endif
    init(maxn);
    int _,n,m;for(scanf("%d",&_);_;_--) {
        scanf("%d%d",&n,&m);
        printf("%lld\n",solve(n,m));
    }
    return 0;
}